What I did is: $f$ has zero at $0$, so $f(z)=zg(z)$.
Using the data, we can say $\frac{g(z)}{e^z}$ is constant.
So, $f(z)=Cze^z$. Is that correct?
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1What does “using the data” mean? – Thomas Andrews Apr 12 '21 at 16:48
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And not all $C$ will work. – Thomas Andrews Apr 12 '21 at 16:50
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$\frac{zg(z)}{ze^z}\leq 1$, So $\frac{g(z)}{e^z}\leq 1$, So $\frac{g(z)}{e^z}$ is a constant function. – Madhan Kumar Apr 12 '21 at 16:51
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why are you saying it? – Madhan Kumar Apr 12 '21 at 16:52
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1If $C=2$ then $|f(z)|=2|ze^z|>|ze^z|$ – morrowmh Apr 12 '21 at 16:53
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That is in no way “using data.” Your answer should include that comment, not the vague “using the data.” State which theorem applies. – Thomas Andrews Apr 12 '21 at 16:54
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1Note that this is a special case of https://math.stackexchange.com/q/52121/42969. – Martin R Apr 12 '21 at 16:57
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Again, clarity. “Why are you saying it?” is unclear. What is “it.” – Thomas Andrews Apr 12 '21 at 16:57
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Doubts cleared @ThomasAndrews. – Madhan Kumar Apr 12 '21 at 17:01
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Fleshing out the steps: substituting $z=0$ gives $|f(0)|\le0$ so $f(0)=0$; since $f$ is entire (Taylor series expansion valid over all complex plane), $f(z)=zg(z)$ for another entire function $g$. Then for $z\ne0$: $$|f(z)|=|z||g(z)|\le|z||e^z|$$ $$|g(z)/e^z|=|g(z)e^{-z}|\le1$$ and $g(z)/e^z$ is entire bounded, so constant by Liouville's theorem. Hence $f(z)=cze^z$; the original condition forces $|c|\le1$.
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